Hello everybody!
I am currently trying to know how much memory of my S32K144 is being used by an application that I made in Simulink.
Looking in the .map file, I find this information:
As I understand, the RAM (m_data) consumption would be of 32 768 B? How is that even possible? Am I reading this the wrong way?
Kind regards,
Juliana
Solved! Go to Solution.
Hello @juliana_rojas30 ,
I think the 'src' is the problem in your path, please open in explorer.
The full path for me is 'C:\AddOnR2019b\Toolboxes\NXP_MBDToolbox_S32K1xx\tools\gcc-6.3-arm32-eabi\bin', but obviously it differs from PC to PC.
Hope this helps,
Marius
Hello @juliana_rojas30 ,
If you want to get the size of the generated code, inside our toolbox, please go to this location {Toolbox installation}\src\tools\gcc-6.3-arm32-eabi\bin there is an executable called arm-none-eabi-size.exe
If you run this executable passing as argument the generated elf you will see the following output:
C:\.....\>arm-none-eabi-size.exe generated.elf
text data bss dec hex filename
46884 1272 6104 54260 d3f4 generated.elf
Hope this helps,
Marius
Hello @mariuslucianand ,
Thanks for your reply. I was trying to look for the direction you said inside the project folder I am working on, but I can't find the folder "tools". There's one called "clock" and another one called "power", but no tools. Actually if I put only the direction directly on the path, I get this:
Should I run any command to generate this folder or something?
Thank you!
Kind regards,
Juliana.
Hello @juliana_rojas30 ,
I think the 'src' is the problem in your path, please open in explorer.
The full path for me is 'C:\AddOnR2019b\Toolboxes\NXP_MBDToolbox_S32K1xx\tools\gcc-6.3-arm32-eabi\bin', but obviously it differs from PC to PC.
Hope this helps,
Marius
hello, when using iMXRT the map file is not getting generated, is there a way to see the ram occupancy (peak ram usage)? i have tried to experiment with the stack profiler from Simulink but i got some errors
It worked perfectly! Thanks for your help!
Kind regards,
Juliana.